October 2006: The locker room (solution)

The state of locker n is changed when the kth student passes through, for every divisor k of n. Since factors usually come in pairs {j, k} where j x k = n, the net effect of students j and k on this locker is nil. The exception is when n is a perfect square, in which case there is no other divisor to cancel the effect of the sqrt(n) student. Therefore, the lockers which are open at the end are exactly the perfect squares, 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

Correct solutions: Evan Templeton, Matt Bachmann, Kaloyan Todorov, Theresa Sparacio ('04), Kristin Jekielek, Hiro Arai, Robert Pehlman, Jenny Witzeling, Dulguun Bayasgalan, Ritoban Basu-Thakur, Ritwik Niyogi, Sunil Baidar, Wade Robertson, Mike Scanish ('01), Lisa Dubbs, Amanda Janiec, Ben Raffeto